Question

How do you find the angle between the vectors `u=3i+4j` and `v=-7i+5j`?

Answer 1

91.33 degrees

Explanation:

We're working in `RR^2` here. The dot product of two vectors given by `vec(u) = ((u_1),(u_2))` and `vec(v) = ((v_1),(v_2))` is given by:

`vec(u)*vec(v) = u_1v_1 + u_2v_2 = |vec(u)||vec(v)|costheta` where theta is the angle between the vectors.

`therefore theta = arccos[(u_1v_1 + u_2v_2)/(|vec(u)||vec(v)|)]`

`|vec(u)| = sqrt(3^2+4^2) = 5`

`|vec(v)| = sqrt((-7)^2 + 5^2) = sqrt(74)`

`theta = arccos[((3)(-7) + (4)(5))/(5sqrt(74))] = 91.33` degrees